C1 Cubics.

This space is of great interest for applications because it
provides the smallest polynomial degree for which
the dimnesnion of S always exceeds the number of
vertices in the triangulation. Thus one can hope to be able
to interpolate to given function values at the vertices of
the triangulation.

If you can prove this statement (or find a counterexample,
perhaps with the MDS applet above) you will be instantly famous,
at least among the people who care about multivariate spline
spaces!

So what's the difficulty? It's hard to localize things.
All kinds of apparently obvious induction arguments fail
because whatever you do in one triangle affects what happens
everywhere else. The main difficulty seems to be that
adding triangles can actually decrease the dimension of
S. For example, adding a fill, i.e., joining a new
triangle on two edges to the triangulation, increases
VI by 1 and decreasesVB by 1. Thus the conjectured dimension decreases by 1.

I have recently been looking into the possibility of using
techniques for proving (**) that were developed for the
solution of the famous Four Color Map problem. You can
examine slides of a recent talk
on this connection.

Another famous open question is

Is it always possible to find a function in S
that assumes given values at the vertices of the
triangulation?

In other words, is it possible for all triangulations and
all sets of function values to interpolate at the vertices of the
triangulation? At first it seems obvious that this must be
possible since the dimension of S exceeds twice the
number of vertices. But it is conceivable, even though no such
case is known, that for some triangulations the
interpolation conditions and the smoothness conditions are
inconsistent.

One way of approaching a proof of (**) is to be a little
more specific and try to prove that one can always set 3
coefficients at and around a boundary vertex, 2 at and
around an interior vertex, and then determine the whole
space by imposing one more point, e.g., the central control
point of a particular triangle. The advantage of making the
stronger statement is that for an induction proof one has a
stronger induction hypothesis to work with. The Figures on
this page nearby show several attempts in that direction,
with blue domain points around boundary vertices, green
points around interior vertices, and the final point (and
the points implied by it) being red.

It turns out that this approach works on the Morgan-Scott
split, and on a generic hexagon, but not on a regular
hexagon, as indicated by the Figures! On the regular
hexagon three points are imposed on every boundary vertex
except the one on the very left. In the 1-disk around that
vertex only two points can be imposed (after imposing three
points around each of the other boundary vertices).

Here is summary of what is and is not known about S:

The lower bound (*) holds.

The equality (**) holds generically, as shown by Billera and Whiteley.
This means that if
it does not hold then an arbitrarily small change in the
location of the vertices will force it to hold.

Billera also showed that if there is a triangulation
for which (**) does not
hold, then the dimension is also non-generic in the case
that d=2.

In the case that d>3 (and r=1) )
singular vertices are the only configurations for which
the dimension of S is not generic.

While the dimesnion is known generically, it is not
known whether the above stated interpolation problem
can always be solved even just generically.